CHE202 Structure & Reactivity in Organic Chemistry: Reduction Reactions and Heterocyclic Chemistry 9 lectures, Semester B 2014

Dr. Chris Jones [email protected]

Office: 1.07 Joseph Priestley Building Office hours: 9.30-10.30 am Tuesday 1.30-2.30 pm Thursday (by appointment only) Course structure and recommended texts 2

§ Coursework: Semester B – week 9 5% (‘Coursework 7’) Semester B – week 11 5% (‘Coursework 8’) § Test: Semester B – week 12 15% (‘Test 4’)

§ Recommended text books:

‘Organic Chemistry’, Clayden, ‘Oxidation & Reduction in ‘Heterocyclic Chemistry’, Greeves & Warren, OUP, 2012. Organic Chemistry’, Donohoe, Joule & Mills, Wiley, 2010. OUP, 2000.

Don’t forget clickers Overview of Reduction Chemistry lecture material 3

§ Reduction: - Definition (recap.) - Reduction of carbon-carbon double and triple bonds - Heterogeneous hydrogenation - Homogeneous hydrogenation, including stereoselective hydrogenation - Dissolved metal reductions - Other methods of reduction

- Reduction of carbon-heteroatom double and triple bonds - Reduction of carbonyl derivatives, addressing chemoselectivity - Stereoselective reduction of carbonyl derivatives - Reduction of imines and nitriles

- Reductive cleavage reactions - Hydrogenolysis of benzyl and allyl groups - Dissolved metal reduction - Deoxygenation reactions

- Reduction of heteroatom functional groups e.g. azides, nitro groups, N-O bond cleavage Reduction: definition 4

§ Reduction of an organic substrate can be defined as:

- The concerted addition of hydrogen. e.g. O H2 (g) H O H Catalytic hydrogenation (e.g. H2(g) & Pd)

- The ionic addition of hydrogen NB: Dr. Lebrasseur’s & Dr Bray’s carbonyl lectures. e.g. O "H " then H Hydride addition then protonation H O H (e.g. LiAlH4 then acid w/up)

- The addition of electrons. e.g. H H electron transfer Dissolved alkali metals

H H one electron added ‘OIL RIG’: a helpful mnemonic... 5

§ Consider the reaction from the point of view of the electrons:

OIL !Oxidation Is Loss

RIG !Reduction Is Gain Question: reduction or not? 6

§ Which of these transformations represent reductions? A. All of them 1 B. 3, 4 and 5 C. 1, 3, 4 and 5 D. 1, 3 and 5 E. 1, 4 and 5 NO2 NO2 NO2 NH2

2 3

N N N N O O O

N 4 NH2

CO Me CO Me 2 5 2 § Reduction of carbon-carbon double and triple bonds Reduction of carbon-carbon double and triple bonds 8

§ Catalytic hydrogenation - Concerted addition of hydrogen across a π-bond. - Use hydrogen gas. - Transition metal (TM) catalyst promotes the reaction. - Catalyst can be heterogeneous or homogeneous.

H H Metal Catalyst H

H2 (g) H

- Hydrogenation has a different mechanism of reduction compared to hydride reducing agents (e.g. NaBH4), therefore different chemoselectivity is often observed. e.g.

O O Pd/C (10 mol%) H H H2 (g) Aldehyde not reduced

NB: 10 mol% = 0.1 equivalent Reduction of carbon-carbon double and triple bonds 9

§ Heterogeneous hydrogenation - Catalyst insoluble in reaction medium.

- TM (e.g. Pt, Pd, Rh) adsorbed onto a solid support, typically carbon or alumina (Al2O3).

§ Reduction of alkenes: - Reactions generally proceed at r.t. and 1 atm. H2 pressure, however, reaction rates increase when elevate T and/or P. - Hydrogenation is typically selective for syn-addition. e.g. Ph Ph PtO2 (cat.) (R) Ph Ph (S) H H2 (1 atm.) H (Z)-alkene

PtO2 (cat.) (S) Ph Ph Ph (S) Ph H H2 (1 atm.) H (E)-alkene

- Need a solvent that dissolves sufficient hydrogen (e.g. methanol, ethanol, acetic acid).

NB: Very occasionally the product from anti-addition occurs, see p. 624 OC textbook for an explanation.! Reduction of carbon-carbon double and triple bonds 10

§ Mechanism: - Complex and difficult to study (reaction occurs on metal surface and each catalyst is different). - Working model (explains syn-selectivity):

i). H2 dissociatively adsorbed onto metal surface. ii). Alkene π-bonds coordinated to catalyst surface. iii). Alkene π-bonds adsorbed onto catalyst surface. iv). A hydrogen atom is added sequentially onto both carbons. v). Reduced product can dissociate from catalyst surface. syn-addition R R R R H2 H H

ADDITION ADSORPTION then DISSOCIATION R R R R R R R R H H H H R R H H R R metal catalyst surface COORDINATION ADSORPTION

- Syn-selectivity increases with increased hydrogen pressure. - Reactivity decreases with increased alkene substitution. Reduction of carbon-carbon double and triple bonds 11

§ Complete reduction of aromatic compounds: - Lose aromaticity: more forcing conditions required c.f. isolated alkenes. - Rh, Ru & Pt are most effective catalysts (i.e. Pd less active so use Pd when require chemoselectivity for alkene reduction in presence of aromatic ring). - Carbocyclic and heterocyclic aromatic rings amenable.

HO O HO O OH OH Pt/C (cat.), H (4 atm.) Carbocyclic ring O 2 O NB: carboxylic acid untouched OH AcOH OH

O O NB: increased

H2 pressures

H 10% PtO2 (10 mol%) H2 (5 atm.), HBr, r.t. Heterocyclic ring N NB: syn-addition N of hydrogen 74% Bun Me n Me Bu

(±)-monomorine Question: reduction of carbon-carbon double and triple bonds 14

§ Which of these structures 1 to 5 is the correct reduction product?

A. 1 O B. 2 Pd/C (10 mol%), H2 (1 atm.) ? C. 3 MeOH, r.t. D. 4 E. 5

OH O O

H H H H H H 1 2 3

O O

H H H H 4 5 Reduction of carbon-carbon double and triple bonds 12

§ Reduction of alkynes to alkanes: - Standard heterogeneous hydrogenation results in complete reduction to alkanes. e.g.

H H O Pd/C (10 mol%) Ph OMe Ph 2 moles of H2 added OMe H2 (1 atm.), MeOH H H O

1 mole of H2 added Ph CO2Me

H H

alkene intermediate

- Alkynes to alkenes?

- Require chemoselectivity to differentiate between reduction of alkyne and alkene. - Require control over cis- or trans- geometry. Reduction of carbon-carbon double and triple bonds 13

§ Reduction of alkynes to cis-alkenes - Lindlar’s catalyst: affords cis-alkenes. - Pd is poisoned with Pb and an amine – more active towards alkyne than alkene (NB: alkene reduction is still possible so reactions often require careful monitoring). e.g. Herbert Lindlar

O Pd/CaCO (10 mol%) Ph CO2Me 3 Solid support – CaCO or BaSO Ph 3 4 H2 (1 atm.), quinoline, OMe H H Pb(OAc)4 – deactivates catalyst Pb(OAc)4, EtOAc Quinoline – competitive cis-(Z)-alkene binding to catalyst surface

e.g. O O N

Pd/CaCO3 (10 mo%) O O H2 (1 atm.), quinoline TBSO O Pb(OAc)4, hexane, r.t. TBSO R R O 86%

- Mechanism: two hydrogens added to same face of alkyne (similar to slide 10), giving syn-addition.! Reduction of carbon-carbon double and triple bonds 15

§ Homogeneous hydrogenation: - Metal-ligand complex is soluble in reaction medium. - Phosphines are common ligands (i.e. good donors). Sir Geoffrey Wilkinson (Nobel Prize in Chemistry 1973) e.g. Wilkinson’s catalyst, (Ph3P)3RhCl - Stereospecific syn-addition of hydrogen across alkene. - Less substituted and least sterically hindered double bonds reduced most easily. - Chemoselectivity (i.e. ketones, carboxylic acids, esters, nitriles, ethers and nitro groups all inert to these conditions.

Least hindered alkene

O O (Ph P) RhCl (cat.), H (1 atm.) O 3 3 2 O O benzene/EtOH O 95%

NB: Heterogeneous hydrogenation (e.g. H2, Pd/C) is non-selective and leads to over reduction

NB: This mechanism is beyond the scope of the course – for a full explanation see ‘Ox & Red in Org Synth’ p. 54 Reduction of carbon-carbon double and triple bonds 16

§ Enantioselective hydrogenation: H2 - If metal-ligand complex (MLn) is chiral, then possible to control to which face of alkene H2 is delivered. - Discrimination between enantiotopic faces of alkene MLn + WHICH FACE? can lead to single enantiomer of product.

H2

ANSWER: use chiral ligand. e.g. BINAP is a chiral & bidentate ligand for TM. - Commercially available as both (R)- and (S)- enantiomers (i.e. chose desired product enantiomer).

Bottom face top face

CO2H CO2H PPhRh(I), (R)-BINAP CO2H Rh(I), (S)-BINAP H 2 H PPh NHCOPh 2H2 (1 atm.) NHCOPh H2 (1 atm.) NHCOPh

Very high facial selectivity (98% ee) NB: this is a protected version (S)-BINAP of the amino acid - alanine

- Beyond hydrogenation, homogeneous catalysis is an extremely important area of organic chemistry and you will encounter numerous examples in future lecture courses… Reduction of carbon-carbon double and triple bonds 17

§ Dissolved metal reductions: - Alkali metals (e.g. Na, Li, K) dissolved in liquid ammonia. - ‘Free electrons’ add to low-lying π*-orbitals.

NH3 (l) Consider the solution as a Na Na + e source of ‘free electrons’

§ Reduction of alkene to alkane: - Selectively reduces electron deficient alkenes to alkanes (i.e. no reaction with standard alkenes).

O O i). Li, NH3 (l), t-BuOH (1 equiv), –78 ˚C

ii). NH4Cl

Reduces electron deficient alkene only

NB: Low temperature as NH3 is a gas at T > –33 ˚C. NB: Stoichiometry of alcohol is crucial. Reduction of carbon-carbon double and triple bonds 18

§ Mechanism of alkene reduction: i). Addition of electron to π-bond gives radical anion. ii). Radical anion protonated by 1 equiv. of t-BuOH to give neutral radical. iii). Addition of 2nd electron yields allylic anion. iv). Proton transfer affords enolate (NB: no proton source so enolate stable under these conditions). v). Addition of more acidic proton source, NH4Cl (aq), protonates enolate to give ketone product.

O O OH + e H Ot-Bu

Single electron transfer Single + e electron transfer

O O PROTON OH H TRANSFER

No t-BuOH remaining – enolate

stable until add NH4Cl (aq) Reduction of carbon-carbon double and triple bonds 19

§ Reduction of alkynes to trans-alkenes: - Dissolved metal reduction forms trans-double bond with high levels of stereoselectivity. e.g. c.f. Lindlar catalyst gives cis-alkene

trans-(E)-alkene

Na, NH3 (l), –33 ˚C

+ e H NH2

H H NH2 + e

Vinyl anions sufficiently basic to deprotonate NH3 (c.f. enolate basicity on slide 18)

- Provided it is not electron deficient, product alkene will not be reduced. - Vinyl anions are geometrically unstable and choose (E)-geometry. Question: some pKa revision

§ Put these anions in order of decreasing stability (i.e. most stable first and least stable last).

A. 1, 5, 2, 3, 4 B. 1, 5, 2, 4, 3 C. 5, 1, 2, 3, 4 D. 1, 2, 5, 4, 3 E. 5, 1, 4, 3, 2

O

1 2 3

N

4 5 Reduction of carbon-carbon double and triple bonds 20

§ Partial reduction of aromatic rings: - Birch reduction: dissolved metal reduction of aromatic rings. - Affords non-conjugated diene product. - Similar mechanism: electron transfer, radical anion protonation… e.g. Arthur Birch H H

Na, NH3 (l), t-BuOH, –33 ˚C

H H

+ e H Ot-Bu

H H H H H H H Ot-Bu + e

Pentadienyl anion has the highest electron density at the central position, hence kinetic protonation (at low T) affords regioselectivity for non-conjugated product. Reduction of carbon-carbon double and triple bonds 21

§ Regioselectivity of Birch reduction: - Electron withdrawing groups promote ipso, para reduction. e.g.

CO2Me CO2Me H Na, NH (l), t-BuOH, –78 ˚C 3 ipso, para

H H

+ e H Ot-Bu

CO Me CO2Me 2 CO2Me CO2Me H Ot-Bu + e

- Electron withdrawing groups stabilise electron density at the ipso and para positions. - Protonation favoured at C-4 as leaves a radical stabilised by conjugation with the carbonyl group. Reduction of carbon-carbon double and triple bonds 22

§ Regioselectivity of Birch reduction: - Electron donating groups promote ortho, meta reduction. e.g.

OMe OMe H Na, NH (l), t-BuOH, –78 ˚C 3 H ortho, meta H H

+ e H Ot-Bu

OMe OMe OMe OMe H Ot-Bu + e

- Electron donating groups stabilise electron density at the ortho and meta positions. - Protonation favoured at C-2 as leaves a radical stabilised by conjugation with the methoxy group. Reduction of carbon-carbon double and triple bonds 23

§ Birch reduction of heterocycles: - Heterocycles bearing an electron withdrawing group can also be partially reduced. e.g.

O O

i-PrO O i-PrO O Na, NH3 (l), t-BuOH, –78 ˚C

+ e H Ot-Bu

O O O H Ot-Bu + e i-PrO O i-PrO O i-PrO O

- Heterocycles are electron rich – electron withdrawing group required for reduction. - Electron withdrawing group also stabilises radical anion formation, controls regioselectivity. Reduction of carbon-carbon double and triple bonds 24

§ Application of the Birch reduction in complex synthesis: - Alkene is conjugated to the benzene ring, therefore electron deficient and can be reduced. - Alkene reduced more rapidly than electron rich benzene ring. e.g.

O O O O

K, NH3 (l) H THF, –70 ˚C H H H MeO MeO Question: Birch reduction 25

§ Which of these substrates will be reduced the fastest under Birch conditions (i.e. dissolved metal, NH3 (l), –78 ˚C)?

CN MeO O

1 2 3

4 5

A. 1 B. 2 C. 3 D. 4 E. 5 Reduction of carbon-carbon double and triple bonds 26

§ Diimide reduction of alkynes and alkenes to alkanes: - Concerted syn-addition of hydrogen across π-bond using cis-diimide. - Alkynes are reduced to alkanes (NB: iodoalkynes are an exception and cis-iodoalkenes produced). e.g.

H H N –N2 (g) syn-addition N H H

cis-diimide Irreversible

N2 evolution

- Diimide is unstable, can be generated in situ from oxidation of hydrazine or decarboxylation of potassium azodicarboxylate.

Cu(II) H H Oxidation H2N NH2 O N N 2 NB: trans-diimide most stable isomer, isomerise to reactive cis-diimide KO C with acid catalysis or heat 2 AcOH H H Decarboxylation N N –2 x CO2 N N CO2K

NB: The mechanisms for diimide formation are beyond the scope of the course. Reduction of carbon-carbon double and triple bonds 27

- Order of reactivity: alkynes > terminal or strained alkenes > substituted alkenes. - trans-Alkenes react faster than cis-alkenes. - Will not reduce polarised double bonds (e.g. C=O). e.g.

KO2CN=NCO2K O O NB: O-O σ-bond intact AcOH, CH Cl O 2 2 O

NH2NH2 NB: trans-alkenes reduced in preference to cis-alkene O2, Cu(II) Reduction of carbon-carbon double and triple bonds 28

§ Reduction of propargylic alcohols to trans-allylic alcohols: - Use LiAlH4. - Stereoselectivity due to complexation of reducing agent to oxygen. e.g.

LiAlH4, THF, 70 ˚C OH Me3Si Me3Si OH

(E)-alkene

LiAlH4 deprotonates alcohol then complex stereospecific forms between H AlH3Li protonation of C-Al H2O alkoxide and Lewis bond during w/up H acidic AlH3 H2 (g) evolution

Li H2Al O H2Al O O Li Me Si Me3Si 3 Me3Si Al H H H Li H H

Intramolecular delivery Intramolecular coordination of vinyl anion to Lewis

of hydride acidic Al species locks geometry of alkene Reduction of carbon-carbon double and triple bonds 29

§ Summary: - C-C π-bonds typically reduced by hydrogenation or dissolved alkali metal. - Heterogeneous catalysts of Pd, Pt, Rh, Ru etc. are effective at C-C π-bond hydrogenation (alkene, alkyne, aromatic). - Hydrogenation of C-C π-bond involves stereoselective syn-addition of hydrogen. - Pd is less active and can be used for chemoselective alkene reduction in presence of aromatic ring. - Mechanism of action means hydrogenation can be chemoselective for C-C π-bond over other FGs (e.g. ketone, nitrile, ester, amide etc.). - Homogenous hydrogenation can offer greater regioselectivity w.r.t. heterogeneous catalysis (e.g. alkene reduction more sensitive to steric requirements) and opportunies for enantioselectivity. - cis-Alkenes formed stereoselectively from alkynes using Lindlar’s catalyst.

- trans-Alkenes formed stereoselectively from alkynes using dissolved alkali metal or LiAlH4 (if O atom). - Regioselectivity of Birch reduction determined by substituents (EWG gives ipso, para; EDG gives ortho, meta). - Diimide is an alternative to TM catalysed hydrogenation and dissolved metal reductions, stereoselective syn-addition of hydrogen, chemoselective for C-C π-bonds. § Reduction of carbon-heteroatom double and triple bonds Reduction of carbon-heteroatom double and triple bonds 31

§ Catalytic hydrogenation: - Possible to reduce C-heteroatom π-bonds with H2 (g) and TM catalyst. e.g.

O PtO2 (5 mol%) H OH H2 (g) (1 atm.)

NB: Pt more reactive metal than Pd

- BUT… chemoselectivity is big issue (i.e. many substrates contain C-C π-bonds that are also reduced).

§ Ionic hydrogenation: - C-heteroatom π-bonds are polarised and susceptible to attack by nucleophilic ‘hydride’. - Opportunity to control chemoselectivity.

O NaBH4, MeOH H OH 0 ˚C, 97 %

NB: see Dr Lebrasseur’s & Dr Bray’s lectures on carbonyl chemistry Reduction of carbon-heteroatom double and triple bonds 32

§ Functional groups: - Wide range of FGs, with wide range of electrophilicities. e.g. O O O

R N R NHR' R R' R Cl NITRILE AMIDE KETONE ACID CHLORIDE

O NH O O

R H R H R OH R OR' ALDEHYDE IMINE CARBOXYLIC ACID ESTER

§ Hydride reducing agents: - Wide variety of reducing agents, with different reactivities. e.g.

LiAlH4, NaBH4, BH3, LiBH4, DIBAL, NaCNBH3, Al(Oi-Pr)3.

Achieve chemoselectivity through careful selection of reducing agent for desired FG reduction. Reduction of carbon-heteroatom double and triple bonds

§ Put these functional groups in order of decreasing electrophilicity (i.e. most reactive first and least reactive last)

O NH O

R NHR' R H R R' AMIDE IMINE KETONE O O

R H R OH ALDEHYDE CARBOXYLIC ACID O

R OR' ESTER

A. Aldehyde, Ketone, Imine, Ester, Amide, Acid B. Imine, Ester, Amide, Ketone, Aldehyde, Acid C. Acid, Amide, Ester, Ketone, Aldehyde, Imine D. Imine, Aldehyde, Ketone, Ester, Amide, Acid E. Imine, Aldehyde, Ketone, Ester, Acid, Amide Reduction of carbon-heteroatom double and triple bonds 33

§ Summary of FG reactivity with common reducing agents:

NR'H O O O O O

R H R H R R' R OR' R NHR' R OH IMINIUM ION ALDEHYDE KETONE ESTER AMIDE CARBOXYLIC ACID

decreasing electrophilicity

✓ ✓ ✓ ✓ ✓ slow LiAlH4

✓ ✓ ✓ slow ✗ ✗ NaBH4

✓ ✓ ✓ ✗ ✗ LiBH4 ✓

DIBAL ✓ ✓ ✓ ✓ ✓ ✓ (–78 ˚C) slow (✗) (✗) (✓ aldehyde) (✓ aldehyde) (✓)

BH3 ✓ slow slow slow ✓ ✓

✓ slow slow ✗ ✗ ✗ NaCNBH3

Product AMINE ALCOHOL ALCOHOL ALCOHOL AMINE ALCOHOL Reduction of carbon-oxygen double and triple bonds 34

§ Reduction of aldehydes and ketones to alcohols (recap.): - Use NaBH4. e.g. O OH

NaBH4, MeOH N N Racemic product H O O

H BH3 H3O w/up NaBH4

MeOH O

N NaBH3OMe + H2 H NB: amide not reduced

MeOH O

NaBH2(OMe)2 + H2

More alkoxy ligands leads to more powerful reducing agent Reduction of carbon-oxygen double and triple bonds 35

§ Reduction of aldehydes and ketones to alcohols (recap.): - Use LiAlH4 (NB: stronger reducing agent than NaBH4 – reduces amide too, no chemoselectivity). e.g. O OH

LiAlH4, THF, ∆ N N H O

LiAlH Reduce amide 4 H3O then w/up

Li O AlH3 O H AlH3 N Li N H O O

NB: mechanism of amide reduction covered later

Li+ functions as a Lewis acid (i.e. precoordination activates the carbonyl by lowering the LUMO energy) Reduction of carbon-oxygen double and triple bonds 36

§ Reduction of aldehydes and ketones to alcohols (alternative): - Meerwein-Ponndorf-Verley reduction. - High chemoselectivity for aldehydes and ketones over esters, alkenes etc. - Works via hydride transfer from the isopropoxy group (i.e. different mechanism c.f. NaBH4 & LiAlH4). e.g.

H H O OH Al(Oi-Pr)3, i-PrOH, ∆ O + MeO2C MeO2C 78 %

MeO2C MeO2C Reaction completely

reversible Al(Oi-Pr) Al is Lewis acid 3 i-PrOH w/up

Oi-Pr O Oi-Pr O Al Oi-Pr Al Oi-Pr hydride transfer H H O O Me Me

6-membered transition state

NB: α,β-unsaturated carbonyls reduced to allylic alcohols Reduction of carbon-oxygen double and triple bonds 37

§ Reduction of α,β-unsaturated ketones: - Two potential sites of attack leads to two possible products. - [1,2]-attack leads to allylic alcohols. - [1,4]-attack leads to fully reduced alcohol (i.e. via conjugate addition, enol tautomerism, reduction). e.g.

O OH OH [1,2]-attack NaBH4 59 : 41 without CeCl3 + what is the ratio? MeOH 99 : 1 with CeCl3

H BH H BH3 3 [1,4]-attack O O

tautomerisation

- Luche reduction: - Excellent regioselectivity for [1,2]-attack by using CeCl3 in addition to NaBH4. - Ce(III) activates carbonyl & promotes formation of alkoxyborohydrides from NaBH4 and MeOH. Question:

§ Identify which statements are correct (more than one statement may be correct).

A. At room temperature LiAlH4 is a solid

B. At room temperature LiAlH4 is a liquid

C. At room temperature NH3 is a liquid

D. At room temperature NH3 is a gas

E. NaBH4 will not reduce an ester

F. NaBH4 will reduce an ester Reduction of carbon-oxygen double and triple bonds 38

§ Reduction of esters to alcohols: - Use LiAlH4 (or LiBH4 for a milder & chemoselective alternative; NB: NaBH4 not reactive enough). e.g.

O

LiAlH4, THF, 0 ˚C O OH

Li+ activates aldehyde reduced LiAlH4 LiAlH4 carbonyl group to alcohol Li H3Al aluminate Li O O O H AlH3 –EtOAlH3 O O H H

tetrahedral intermediate aldehyde intermediate collapses reduced rapidly

NB: DIBAL at r.t. will reduce esters to alcohols

NB: aluminate species eliminated, but are poorer hydride donors than parent LiAlH4 (c.f. alkoxyborates)

Can we stop the reduction at the aldehyde? Reduction of carbon-oxygen double and triple bonds 39

§ Reduction of esters to aldehydes: - Require tetrahedral intermediate not to collapse to aldehyde in situ. - Use 1 equivalent of DIBAL at low temperature. - DIBAL forms Lewis acid-base complex with carbonyl group in order to become a reducing agent (i.e. reduces more electron rich C=O groups more quickly). e.g.

NB: alkene not reduced O DIBAL, Toluene, –78 ˚C O O

H i-Bu Al is Lewis acid Al H3O i-Bu

i-Bu i-Bu i-Bu OH Al Al H3O H O i-Bu O O H O O H hemiacetal

DIBAL reduces tetrahedral intermediate

esters at –70 ˚C stabilised at low T!

Intermediate collapses to aldehyde upon w/up, but excess DIBAL has already been quenched Reduction of carbon-oxygen double and triple bonds 40

§ Reduction of amides to amines: - Amides poor electrophiles, require LiAlH4. e.g.

O

LiAlH4, Et2O NMe2 NMe 2 88 %

Li+ activates LiAlH4 H AlH carbonyl group 3

Li Li H3Al O O H H AlH3 –H3AlOLi

NMe2 NMe2 NMe2 H

tetrahedral intermediate iminium ion reduced

collapses by expelling best rapidly by LiAlH4 LG (i.e. O better than N) Reduction of carbon-oxygen double and triple bonds 41

§ Reduction of amides to amines (alternative):

- Use BH3 (chemoselective for amide over ester). NB: ester not reduced e.g. MeO C MeO2C BH3•THF 2 O N H N H B O Bn Bn H Like DIBAL, BH H H 3 B Borane (B H ) is a gas; forms Lewis acid- 2 6 stabilised as a liquid in H3O w/up base complex H BH3 complex with THF H H MeO2C H MeO2C B H H O N N BH2 Bn Bn

Borate donates BH3 reduces hydride to reactive

reactive imidate iminium ion H MeO2C MeO2C H B H –BH2O H O N H N Bn H B Bn 2 Reduction of carbon-oxygen double and triple bonds 42

§ Reduction of amides to aldehydes? - Require stabilised tetrahedral intermediate following hydride addition (c.f. partial reduction of ester). e.g.

O M-[H] O NMe2

M-[H] hydrolysis to aldehyde

M M O O OH [H] H3O

NMe2 NMe2 NMe2 H

stabilise tetrahedral w/up reveals

intermediate? hemiaminal

- Tetrahedral intermediate is difficult to stabilise and collapses in situ to iminium ion which is further reduced to amine (c.f. LiAlH4). - No general reagent for amide reduction to aldehyde… BUT… Reduction of carbon-oxygen double and triple bonds 43

§ Reduction of amides to aldehydes: - Weinreb’s amides form stable chelated intermediates at low T with DIBAL and LiAlH4. e.g. with DIBAL

O OMe DIBAL, toluene O N 0 ˚C, 74 %

i-Bu OMe Al is Lewis acid H Al – HN i-Bu

hydrolysis to aldehyde i-Bu i-Bu i-Bu Al Al i-Bu H O O OH H O OMe OMe 3 OMe N N N H H

Borate donates hydride stabilised chelated w/up reveals

to reactive imidate tetrahedral intermediate! hemiaminal Question: apply this understanding to a new situation 44

§ Weinreb amides and organolithium reagents (a quick aside): - Considering the previous slide, determine the product of the following reaction: e.g.

O O OMe MeLi, THF ketone N ? Me 0 ˚C

OMe NB: also works with Me Li – HN elimination of

Grignard reagents MeNHOMe

Li O OH OMe H3O OMe N N Me Me

stabilised chelated

tetrahedral intermediate Reduction of carbon-oxygen double and triple bonds 45

§ Reduction of carboxylic acids to alcohols: 1). Use LiAlH4 but require very forcing conditions, as form unreactive carboxylate salts. e.g. 1

O O

LiAlH4 heat OH OH O Li –H2 (g)

unreactive

2). Use BH3: Lewis acidic, so chemoselective for most electron rich carbonyl groups (acids and amides). - Mechanism: complexation with lone pair forms active reducing agent (see amide reduction).

- Milder conditions w.r.t. LiAlH4. e.g. 2 OH

BH3•THF OH O

NB: for further explanation see ‘OC’ textbook p. 619. Reduction of carbon-oxygen double and triple bonds 46

3). Via more reactive intermediate: convert carboxylic acid to mixed anhydride, then reduce with NaBH4. - 2 steps, but generally quick and high yielding procedures. - Chemoselective reduction of acid in presence of ester. e.g. 3 acid alcohol

O O O O BocHN BocHN Cl OEt BocHN NaBH4 OH OH O OEt MeOH, 0 ˚C Et3N, CH2Cl2 97 % yield over 2 steps mixed anhydride H BH3 NaBH4 NB: most reactive at C=O that originated from carboxylic acid, O as other C=O π-system has O O overlap from two oxygens. –CO2 (g) BocHN BocHN H O OEt H –EtO

- Eliminate CO2 (g) and EtO Reduction of carbon-nitrogen double and triple bonds 47

§ Reduction of imines to amines: - Use NaBH4 or LiAlH4. e.g.

N NaBH4, MeOH HN NB: analogous to mechanism 0 ˚C, 95 % H of aldehyde reduction

§ Reductive amination: - Convert carbonyl group to amine through in situ imine formation. - Add acid to increase imine reactivity w.r.t. carbonyl group, can now use much weaker hydride

reducing agent (e.g. NaCNBH3) to chemoselectively reduce iminium ion in presence of carbonyl. e.g.

H O NBn NHBn BnNH2 H BH2NaCN H MeOH, HCl H

Aldehyde not reduced iminium ion NB: can also use

by NaCNBH3 (highly reactive) NaBH(OAc)3 Question: reductive amination 48

§ Eschweiler-Clark reaction: - Introduced in 1st year. - Can you draw the mechanism for this reductive amination? e.g.

N (CH2O)n (aq.) N H HCO2H

O O

H H H O CO2 (g) evolution

N

NB: iminium ion formation H H Reduction of carbon-nitrogen double and triple bonds 49

§ Reduction of nitriles to amines: - Complete reduction using LiAlH4 (i.e. powerful reducing agent). e.g.

NH2 LiAlH4, THF N

reduction of Li+ activates LiAlH4, then LiAlH4 intermediate nitrile group imine followed H3O by w/up

H AlH N 3 Li N Li H

Overall 2 equivalents of hydride added Reduction of carbon-nitrogen double and triple bonds 50

§ Reduction of nitriles to aldehydes: - Partial reduction using DIBAL (1 equiv.). e.g.

O DIBAL (1 equiv.), Toluene, –78 ˚C N H then H3O

i-Bu H Al H3O w/up i-Bu

N i-Bu N i-Bu Al Al H i-Bu H i-Bu

iminoalane H N Al i-Bu i-Bu

NB: Lewis acidic DIBAL coordinates to nitrile lone pair and ‘ate’ complex is hydride donor Reduction of carbon-heteroatom double and triple bonds 51

§ Summary of FG reactivity with common reducing agents:

NR'H O O O O O

R H R H R R' R OR' R NHR' R OH IMINIUM ION ALDEHYDE KETONE ESTER AMIDE CARBOXYLIC ACID

decreasing electrophilicity

✓ ✓ ✓ ✓ ✓ slow LiAlH4

✓ ✓ ✓ slow ✗ ✗ NaBH4

✓ ✓ ✓ ✗ ✗ LiBH4 ✓

DIBAL ✓ ✓ ✓ ✓ ✓ ✓ (–78 ˚C) slow (✗) (✗) (✓ aldehyde) (✓ aldehyde) (✓)

BH3 ✓ slow slow slow ✓ ✓

✓ slow slow ✗ ✗ ✗ NaCNBH3

Product AMINE ALCOHOL ALCOHOL ALCOHOL AMINE ALCOHOL § Stereoselective reduction of carbonyl groups Diastereoselectivity with hydrides: 1,2-stereoinduction 53

§ Felkin-Anh model (recap.): NB: see Dr Bray’s 1st year lectures notes e.g.

O OH LiBH(s-Bu) Ph 3 Ph THF

‘Felkin’ product

- Drawing Newman projections (recap.).

O O O add substituents Me Ph Ph H H Me

Look down highlighted Convert to Newman Ensure configuration

C-C bond projection around chiral centre is correct Diastereoselectivity with hydrides: 1,2-stereoinduction 53

§ Felkin-Anh model (recap.): e.g.

Lowest energy conformations of

chiral starting material

O OH O O LiBH(s-Bu) Me H Ph 3 Ph Ph or Ph THF H Me

‘Felkin’ product (i.e. place largest group 90˚ to C=O)

O - Reactive confirmation (i.e. least sterically hindered). Me … - Nu approaches along Bürgi-Dunitz trajectory (107˚ O=C Nu). Ph R B - Least hindered approach is past ‘H’ and away from large 3 H group (e.g. Ph). H

Newman projection

NB: see Dr Bray’s 1st year lectures notes Diastereoselectivity with hydrides: 1,2-stereoinduction 54

§ Felkin polar model: - Use when α-substituent is electron withdrawing but not a good LG (NB: competing SN2 reaction). e.g.

O O OH i-Pr LiBH(s-Bu) MeS 3 BR3 i-Pr i-Pr H THF H SMe SMe

‘anti-Felkin’ product Newman projection

π* C=O overlaps with σ* C-S;

creates lower energy LUMO

- Reactive conformation (i.e. π* and σ* combine C-S σ* to lower LUMO). SMe SMe - Electronegative group 90˚ to C=O group (even combine if it isn’t the largest group). i-Pr i-Pr H O H O - Approach along Bürgi-Dunitz trajectory (107˚). - Least hindered approach is past ‘H’ and away C=O π* new LUMO from electronegative group. (π* + σ*)

(i.e. most reactive in this conformation) Diastereoselectivity with hydrides: 1,2-stereoinduction 55

§ Felkin chelation model: - Can reverse selectivity when (i) α-substituent contains lone pairs and (ii) use chelating metal. - Require Lewis acid metal that chelates to more than one heteroatom at once (i.e. to the carbonyl group and α-substituent). e.g. Chelating metals

Li+ (sometimes) O OH Mg2+ Zn(BH ) Me 4 2 Me Zn2+ THF Cu2+ Ti4+ OMe OMe Ce3+ Mn2+ ‘Felkin’ product

+ + NB: Na and K do not chelate

- Reactive conformation (i.e. Lewis acid coordination lowers LUMO). Zn2+ Electronegative group almost eclipses C=O to enable - O chelation with the metal (even if electronegative group is MeO Me the largest). R B 3 H - Approach along Bürgi-Dunitz trajectory (107˚). H - Least hindered approach is past ‘H’ and away from group at 90˚. Newman projection Diastereoselectivity with hydrides: 1,2-stereoinduction 56

§ Examples: e.g. 1 Key step in synthesis of anticancer agent, dolastatin.

O

NBn2 NBn2 NBn2 MeO H + CO2Me CO2Me Aldol reaction O OH OH

1 diastereoselectivity 20

- How do we explain diastereoselectivity? Consider reactive conformation. O H CO2Me 1. No chelation so electronegative group goes 90˚ to C=O 2. Approach alongside H H NBn2

O H O NBn O 2 Bn N O 2 Bn2N H H H H H CO2Me OMe Visualisation aid: draw product in most reactive conformation same orientation, then rotate to put longest chains in same plane Diastereoselectivity with hydrides: 1,2-stereoinduction 57

§ Examples: e.g. 2 Chelation or not?

O HO Me HO Me Me2Mg Ph + Ph Ph THF, –70 ˚C OR OR OMe

diastereoselectivity

R = Me 99 1

R = OSi(i-Pr)3 42 58

- How do we explain diastereoselectivity? Consider reactive conformation.

‘R’ = Me ‘R’ = OSi(i-Pr)3

Mg2+ O O Me RO RO 1 Large group disrupts efficient chelation Me H Nu H Ph Nu Ph

Chelation model Felkin polar model Diastereoselectivity with hydrides: 1,2-stereoinduction 58

§ Summary: which model to use?

O Is there a heteroatom at the chiral centre? X R Y Z No Yes α-chiral carbonyl

compound Use Felkin-Anh model Is there a metal (i.e. conformations with the ion capable of

largest group 90˚ to C=O) chelation?

No

Yes

Use Felkin polar model Use Felkin chelation (i.e. conformations with model (i.e. conformations the electronegative with the heteroatom and

group 90˚ to C=O) C=O almost eclipsed) Aims: at the end of this section you will be able to…

§ Understand the origin and predict the stereochemistry of reductions that form 1,3-diols.

§ Understand the origin and predict the stereochemistry of reductions of cyclohexanones (i.e. axial or equatorial attack?).

§ Understand the origin and predict the stereochemistry of the enantioselective CBS reduction of ketones.

§ Understand the term ‘reductive cleavage’ and be familiar with suitable reagents and conditions.

§ Understand the different mechanisms and methods to reduce heteroatom functional groups. Diastereoselectivity with hydrides: 1,3-stereoinduction 59

§ 1,3-Polyols: - Important motif found in many natural products.

1,3-syn 1,3-anti

OH OH OH OH OH

O HO

O OH

(+)-Roxaticin

- How do we control diastereoselectivity?

OH O [H] OH OH OH OH +

1,3-syn 1,3-anti Diastereoselectivity with hydrides: 1,3-stereoinduction 59

§ Conformations of 6-membered rings (recap.): - Cyclohexane is not flat… puckered to enable tetrahedral carbon atoms. - Two most common conformers are the ‘chair’ (lowest energy) and ‘boat’. e.g. NB: axial H H H H H H H NB: equatorial H H H

Cyclohexane Chair Boat

- Cyclohexene adopts a ‘half chair’ in its lowest energy conformation.

H H H H H H H H

Cyclohexene Half chair

NB: for further explanation see ‘OC’ textbook p. 370-374. Diastereoselectivity with hydrides: 1,3-stereoinduction 60

§ 1,3-syn Diols: - Use a Lewis acid (e.g. Bu2BOMe, BF3). - Chelates to carbonyl and alcohol in a 6-membered ring (i.e. half chair conformation). - Favours intermolecular hydride delivery from less hindered face of C=O.

half chair chair

Bu Bu Bu Bu B B OH O Bu BOMe, MeOH O O O O then H O OH OH 2 H 3 NaH B H i-Pr 3 i-Pr i-Pr w/up i-Pr

Lewis acidic boron: 6-membered Why does hydride 1,3-syn (i) chelates C=O and C-OH ring chelate attack from this (ii) activates C=O group face?...consider in 3D

axial attack! NaH3B H H Bu Bu Axial attack Me H Me B ensures smallest O B O Bu group is in axial i-Pr Bu O O position on chair i-Pr

half chair chair Diastereoselectivity with hydrides: 1,3-stereoinduction 61

§ 1,3-anti Diols: - Use weakly Lewis acidic reducing agent to disfavour chelation. - Intramolecular delivery of hydride. - Use weak reducing agent to minimise competition from intermolecular delivery. - Chair-like 6-membered ring T.S. controls stereochemistry.

Intramolecular hydride

delivery

H OAc OH O Me NBH(OAc) O OH OH 4 3 B O OAc i-Pr AcOH H i-Pr i-Pr

Lewis acidic boron chelates Place substituents in lowest 1,3-anti C-OH group energy pseudo-equatorial positions within chair-like 6-

AcOH solvent activates membered ring T.S. C=O group Diastereoselectivity with hydrides: addition to cyclohexanones 62

§ Axial or equatorial attack? - When reducing cyclohexanones, the hydride can end up in an axial or equatorial position. - Can we achieve this selectively?

more hindered approach

H H H H H OH O H 'AXIAL H 'EQUATORIAL H OH ATTACK' ATTACK' H Small hydride Bulky hydride H H H H H H

H

less hindered approach e.g. LiAlH4 e.g. Na(s-Bu)3BH 9:1 ax:eq 4:96 ax:eq

- ‘Axial attack’ favoured with a small hydride source (e.g. LiAlH4, NaBH4).

- ‘Equatorial attack’ favoured by bulky hydride sources (e.g. L-selectride Na(s-Bu)3BH). Diastereoselectivity with hydrides: addition to cyclohexanones 63

§ Why does ‘axial attack’ occur at all? - If ‘axial attack’ is more hindered then why is it even favoured with small hydride sources? - Need to consider the transition state leading to the alkoxy intermediate from ‘axial attack’:

(i.e. more hindered approach)

O 'AXIAL ATTACK'

H H oxy substituent H rotates H H O O H H

'EQUATORIAL alkoxy intermediate ATTACK' from ‘axial attack’

(i.e. less hindered approach)

- ‘Axial attack’: oxy substituent moves away from neighbouring C-H bond. - ‘Equatorial attack’: oxy substituent moves towards neighbouring C-H bond, leading to higher torsional strain in T.S (i.e. disfavoured).

NB: for further explanation see ‘OC’ textbook p. 471. Question: reduction of cyclohexanones

§ What is the stereochemistry of the product of the following reduction? O A. 1

NaBH4 B. 2 ? BnO MeOH C. 3 D. 4

OH OH

BnO BnO

OH OH 1 2

BnO BnO

3 4 Enantioselective reduction of ketones 64

§ In general: - Reagent determines from which face of carbonyl group the hydride approaches. - Selective synthesis of one enantiomer of secondary alcohol over the other. - Require a chiral reagent in order to introduce a chiral environment for the reduction. - Highly desirable but difficult to achieve in practice.

Hydride approaches from top face

H OH R R R or R' O R' R' OH H

top face bottom face

Hydride approaches from bottom face

- Can the reagent be used in catalytic amounts (i.e. chiral and non-racemic compounds are often more expensive than racemic mixtures, so want to use as little as possible)? Enantioselective reduction of ketones 64

§ Corey-Bakshi-Shibata (‘CBS’) reduction: - Uses a chiral reducing agent.

- CBS reagent used in catalytic quantities (also need BH3 as the stoichiometric source of hydride). - Reduce unsymmetrical ketones to chiral secondary alcohols.

- Catalyst binds to both BH3 and substrate in an ordered manner, resulting in high enantioselectivity.

H Ph Boron in catalyst is Lewis acidic and activates C=O group N Ph BH3 used to regenerate active catalyst H3B B O Boat-like arrangement controls facial selectivity Me

‘CBS’ reagent with Binding of BH3 and ketone to exo face of catalyst

BH3 bound

Ph O Me O 10 mol% catalyst B OH Ph N Me BH , THF Ph Me 3 B O Ph Me H H H Ph

Hydride delivered from ‘top’ face (R)-alcohol, 97 % ee

NB: for further explanation see ‘Ox and Red’ textbook p. 69. Enantioselective reduction of ketones 65

§ Predict the stereochemistry of the products of these ‘CBS’ reductions: e.g. 1 H smaller substituent Ph pseudo axial N Ph B O Ph O Me O Me 1 mol% B OH N Cl Cl Ph ? BH , THF Cl Ph 3 B O Ph H H H Ph 96 % ee

e.g. 2 H Ph N Ph B O O Ph O Me OH Me 1 mol% B TBSO Ph ?N Me TBSO Me BH , THF Me 3 B O H H H 94 % de

NB: require good size differentiation between two ketone substituents for high levels of enantioselectivity. § Reductive cleavage reactions Reductive cleavage reactions 67

§ Definition: Break single bonds between carbon and electronegative elements and replace with bonds to hydrogen. REDUCTION C X C H X = N, O, S, halogen

§ Hydrogenolysis - Reductive cleavage of a carbon-heteroatom (C-X) single bond through the addition of hydrogen. - Most commonly used to cleave benzylic and allylic groups from oxygen and nitrogen. - Pd is usual choice of metal catalyst as it reduces the C-X bond faster than the aromatic ring π- bonds. (c.f. Pt, Rh, Ru – see slide 9). e.g.

Ph Pd/C (10 mol%), H2 (1 atm.) N NH EtOH

O O

Pd/C (10 mol%), H2 (1 atm.) O Ph OH EtOAc

Ease of formation and removal means that the benzyl group is commonly used as a ‘protecting group’ in organic synthesis for amines, amides, alcohols and carboxylic acids.

NB: This cleavage mechanism is beyond the scope of the course – for an explanation see ‘Ox & Red in Org Synth’ p. 77 Reductive cleavage reactions 68

- Reductive dehalogenation: - Typically use Pd and a base (NB: produce HX acid as a byproduct which may retard reaction if it isn’t neutralised by base).

- Csp3-Hal and Csp2-Hal bonds are amenable to hydrogenolysis. e.g. Br H OBn OBn Pd/C (10 mol%) Weaker bonds reduced more easily: C-I > C-Br > C-Cl > C-F H2 (1 atm.), Et3N OMe MeOH, r.t., 97% OMe OMe OMe

NB: This mechanism is beyond the scope of the course – for an explanation see ‘Ox & Red in Org Synth’ p. 76

§ Dissolved metal: - An alternative method to cleave N-benzyl and O-benzyl groups. e.g. O Bn O H H H Bn N Na, NH (l), t-BuOH, –78 ˚C N N 3 HN

H H

NB: This mechanism is beyond the scope of the course – for an explanation see ‘Ox & Red in Org Synth’ p. 76 Reductive cleavage reactions 70

§ Deoxygenation – ketones and aldehydes to alkanes: e.g. 1 Reduction to alcohol, make into good LG, displace with hydride.

H H H 1). LiAlH4 OTs 3). H AlH3Li O H 2). TsCl THF

3 steps overall SN2 reaction (i.e. reactivity 1˚ > 2˚ >> 3˚)

- Use Super-Hydride™ for displacement step (i.e. LiEt3BH). - Electron-donating alkyl groups make it the most nucleophilic hydride source.

OMs H neopentyl LiEt3BH, THF Ph Ph

NB: Especially effective for challenging SN2 reactions on activated LG Reductive cleavage reactions 71

e.g. 2 Wolff-Kischner reduction: - Use hydrazine and KOH. - Requires elevated temperatures (up to 200 ˚C).

O H H

NH2NH2, KOH Nicolai Kischner ethylene glycol, ∆

Ethylene glycol is a high boiling point solvent (197 ˚C) NH2NH2 H OR'

H N R N H Hydrazone H R

OH –N2 (g) Nitrogen gas evolution (irreversible) H N NH N N N H OR' N OH R Azine R R H H NB: see ‘Ox & Red in Org Synth’ p. 82 Reductive cleavage reactions: summary 72

§ Which conditions break which bond?

Benzyl group cleavage!

H X Ph X Use Pd/C, H2; or dissolved metal X = O or N

Reductive dehalogenation!

X H C C Use Pd/C, H2

X = Halogen (i.e. I, Br, Cl, F)

Deoxygenation!

O H H Wolf-Kischner or 3 step sequence R R (i.e. reduction, activation, reduction) Question

§ Which set of reagents and conditions A to E will deliver the reduction product?

O ? O HO OH OBn HO

A. H2 (1 atm.), Pd/C, MeOH, r.t.

B. H2 (1 atm.), Pd/BaSO4, Pb(OAc)4, quinoline, r.t.

C. Na, NH3, r.t.

D. Li, NH3, –78 ˚C

E. H2 (1 atm.), Pd/C, MeOH, 50 ˚C § Reduction of heteroatom functional groups Reduction of nitrogen-containing functional groups 75

§ Reduction of azides (-N3) to amines (-NH2): 1). Use H2 & Pd/C to reduce azide (NB: if alkene also present in substrate then use Lindlar’s catalyst as does not reduce alkenes, chemoselectivity). e.g.

alkene intact

Pd/BaSO4 NaN H (1 atm.) OTs 3 N3 2 NH2

DMF Pb(OAc)4

SN2 azide amine

- Strategy: azide is excellent nucleophile, especially for SN2 (i.e. charged, small size, low basicity), so good method of introducing amine into a molecule is azide displacement reaction then reduction. Reduction of nitrogen-containing functional groups 75

2). Staudinger reduction: - use PPh3 as another chemoselective method for azide reduction in the presence of an alkene. e.g.

PPh3, THF

N3 H2N

azide amine

NB: Pd/C & H2 would also reduce alkene

NB: This mechanism is beyond the scope of the course, for further details see ‘Ox & Red in Org Synth’ p. 47 Reduction of nitrogen-containing functional groups 76

§ Reduction of nitro groups (-NO2) to amine (-NH2): - H2 & Pd/C is most common method. - Aromatic and aliphatic nitro groups are reduced. e.g.

CF3 CF3

H2 (1 atm.), Pd/C (10 mol%) EtOH, r.t. NO2 NH2 OEt OEt

- Mechanism:

O O O H H H2 addition H2 addition H2 addition Ar N Ar N Ar N Ar N –H O –H O O 2 H 2 H

- Many reagents reduce nitro group (NB: choose conditions for desired chemoselectivity). e.g. Fe & HCl; FeCl3 & H2O; SnCl2 & HCl; Zn & HCl

NB: For details of this mechanism via SET see ‘Ox & Red in Org Synth’ p. 46 Question: reduction as a tool in total synthesis

§ Identify the correct product of this reaction.

O MeO O N N3 N OMe 10% Pd/C, H2 (1 atm.) O ? H N Me 2 MeOH:EtOAc OBn

Br OMe OMe A. 1 B. 2 C. 3

O O O MeO MeO MeO O O O N N N H2N N OMe H2N N OMe N3 N OMe O O O H2N Me H2N Me H2N Me 1 OBn 2 OH 3 OH

Br OMe OMe Br OMe OMe OMe OMe Heterocyclic Chemistry Aims for this session 3

§ Heterocycles: - Definition. - Why heterocycles are important. - Nomenclature of common heterocycles. - Recap. of key reactions of ketones, enols, imines and enamines. What is a heterocycle? 77

§ A cyclic system (or ‘ring’) that contains one or more heteroatoms. - Can be aromatic or non-aromatic. e.g.

CARBOCYCLIC HETEROCYCLIC

H H N N N H

CYCLOPENTANE PYRROLIDINE 2,3-DIHYDROIMIDAZOLE (non-aromatic) (non-aromatic) (non-aromatic)

N N

N

BENZENE PYRIDINE PYRIMIDINE (aromatic) (aromatic) (aromatic)

NB: multiple heteroatoms Why are heterocycles important? 77

§ Heterocycles are an extremely common motif. - Found in almost half of all known organic compounds. - Key constituents of natural products, DNA, amino acid proline (i.e. proteins), medicines, agrochemicals, materials and dyes.

e.g. NH2 N Me N N O N N HO N N HO N MeO Me N O Quinine O Me N Caffeine OH Deoxyadenosine

H N O Me OMe S N MeO N Me Me N N NO2 Nicotine Omeprazole Ranitidine Me Me N Me N S N H H O Question: identify the heterocycles

§ How many heterocycles are in this pharmaceutical compound? (NB: count any fused rings separately)

O Me Me N N O HN N N S N O i-Pr OEt

Sildenafil (‘Viagra’) A. 0 B. 1 C. 2 D. 3 E. 4 F. 5 Top 20 global best-selling drugs in 2009 78

NB: Drugs containing heterocycles indicated in bold Heterocycles: nomenclature 78

§ Five membered rings with one heteroatom: - NB: we will only study the highlighted compounds in more detail. e.g.

H H H N O N O N O

pyrrole furan pyrrolidine tetrahydrofuran pyrrolidinone (‘THF’)

H X N X X

thiophene(X = S) indole benzofuran (X = O) isoindole (X = N) selenophene (X = Se) benzothiophene (X = S) isobenzofuran (X = O) etc. etc. etc. Heterocycles: nomenclature 78

§ Five membered rings with more than one heteroatom: - NB: we will not study these heterocycles in detail, but you should be aware of these compounds. e.g.

H H H H H N N N N N N N N N N N N N N

imidazole pyrazole 1,2,3-triazole 1,2,4-triazole tetrazole

H H H H N N N N N O S O S O

oxazole thiazole isoxazole isothiazole benzoxazole Heterocycles: nomenclature 78

§ Six membered rings with one heteroatom: - NB: we will only study the highlighted compounds in more detail. e.g.

8 N N O 7 1 2 N NB: aromatic 6 3

5 4

pyridine quinoline isoquinoline pyrylium cation

H H H N N N O O 6 1 2 NB: non-aromatic 5 3

4

piperidine 1,2,3,4-tetrahydropyridine pyridone tetrahydropyran Heterocycles: nomenclature 78

§ Six membered rings with more than one heteroatom: - NB: we will not study these heterocycles in detail, but you should be aware of these compounds. e.g.

N N N N N NB: aromatic N N N N

pyridazine pyrimidine 1,3,5-triazine quinazoline

H H N N O NB: non-aromatic N O H O

piperazine morpholine 1,4-dioxane Heterocycles: nomenclature 78

§ Beyond five and six membered rings: - NB: we will only study the highlighted compound in more detail. e.g.

O O NH NH O NH epoxide aziridine oxetane azetidine β-lactam

OH OH OH OH OH H PhO N S Me O O HO N Me O CO H O OH 2

(+)-Roxaticin Penicillins

NB: macrocycle Heterocycles: key reactions (recap.) 78

§ Enol formation: - NB: acidic conditions.

e.g. NB: consider pKa of α-proton

H H O H OH2 O H2O O H protonation H deprotonation R R R

ketone enol

§ Enolate formation: - NB: basic conditions.

e.g. NB: consider pKa of α-proton

Li NB: need a strong base O Ni-Pr2 O to effect complete irreversible deprotonation H R R of a ketone (i.e. use LDA)

ketone enolate

NB: for a comprehensive list of pKa values, see: http://evans.harvard.edu/pdf/evans_pka_table.pdf Question: pKa (again!)

§ What are the pKa values of the two α-protons highlighted below?

H H O H OH2 O H2O O H H R R R

A. They are both the same B. 43 and 26 C. 26 and –6 D. 43 and –6 Aims for this session 3

§ Heterocycles: - Continue recap. of key reactions of ketones, enols, imines and enamines. - Recap. of aromaticity. - Look at pyridine: structure and reactivity (how does it compare to benzene?). - Look at pyrrole: structure and reactivity (how does it compare to benzene?). Heterocycles: key reactions (recap.) 78

§ Formation of imines: - NB: typically require catalytic acid. e.g.

O NR RNH2 + H2O cat. H+ R R

ketone imine

§ Enamine formation: - NB: occurs under acidic or basic conditions. e.g. c.f. enol and enolate formation

R NB: C=N bond weaker NR tautomerisation HN than C=O, therefore easier H to form enamine compared R R to enol or enolate

imine enamine

NB: enamines react with electrophiles on carbon and not nitrogen Heterocycles: key reactions (recap.) 78

§ Aldol: - NB: formation of thermodynamically favoured α,β-unsaturated product drives reaction. e.g.

H O O cat. H+ O + R H Ph R Ph

enol aldehyde α,β-unsaturated product Heterocycles: key reactions (recap.) 78

§ Mannich reaction: - NB: electrophile is imine rather than aldehyde. - Amine is poorer LG (c.f. alcohol in Aldol reaction), therefore intermediate does not eliminate. e.g.

H O NR cat. H+ O NHR + R H Ph R Ph

enol imine amine product

§ Conjugate addition: - NB: orbital-controlled [1,4]-nucleophilic addition. - NB: called a “Michael addition’ when carbon-based nucleophile such as enol or enolate. e.g.

O Nu OH O

R R Nu R Nu

enone enol intermediate

NB: favoured when using ‘soft’ nucleophiles (i.e. uncharged and/or diffuse orbitals) Aromaticity (recap.) 78

§ Hückel’s Rule: e.g.

2π electrons 6π electrons 6π electrons 6π electrons

H N PYRROLE! N PYRIDINE! isoelectronic and isolobal isoelectronic and isolobal

with cyclopentadiene anion with benzene

6π electrons 6π electrons

Planar structures which have a cyclic array of [4n + 2] delocalisable π-electrons Question: heteroaromatic?

§ Identify the heterocycle that is not aromatic

H H N O N N O A B C

H H N N

D E F

A. A B. B C. C D. D E. E F. F Pyridine 105

§ Structure: - Isoelectronic with benzene (i.e. 6π electrons, one in each of 6 parallel p orbitals). e.g.

sp2 hybridised carbon: H one electron in sp2 orbital σ-bonding to H H

benzene one electron in each p orbital

sp2 hybridised nitrogen: N lone pair of electrons in N sp2 orbital, orthogonal to plane of π-system

pyridine one electron in each p orbital

Nitrogen atom has major effect on reactivity and properties of pyridine w.r.t benzene Pyridine: is it really aromatic? 106

§ Resonance energy (kJ/mol): - i.e. energy gained through delocalisation around the ring. e.g.

150 117 i.e. benzene gains more energy through N resonance

§ Bond lengths (pm): - pyridine has different bond lengths within ring system. e.g.

139.4 139.5

i.e. shorter bond lengths in pyridine 140 134 N Pyridine: is it really aromatic? 107

§ 1H NMR spectroscopic chemical shifts (i.e. δH in ppm): - i.e. how deshielded is each environment? e.g.

H 7.1

H 7.5 H H 8.5

i.e. more deshielding in pyridine 7.2 N

§ 13C NMR spectroscopic chemical shifts (i.e. δH in ppm): - i.e. how deshielded is each environment? e.g.

123.6

135.7 149.8 128.5 i.e. more deshielding in pyridine N

YES, pyridine is aromatic Question: hybridisation states in heterocycles

§ In which hybrid atomic orbital is the nitrogen lone pair located?

N

A. p B. sp C. sp2 D. sp3 Question: hybridisation states in heterocycles

§ In which hybrid atomic orbital is the nitrogen lone pair located?

H N

A. p B. sp C. sp2 D. sp3 Pyridine: general reactivity trends 110

§ Lone pair reactivity: - Basic (i.e. can be easily protonated). - Nucleophilic (i.e. can be alkylated or acylated). - N-oxidation (form N-oxides, see later).

O X N E e.g. E = H R X R

protonation alkylation acylation

§ Benzene-like reactivity: - Attack on π-system. E NB: SEAr is less likely than with benzene - Electrophilic substitution. N

§ Imine-like reactivity: - Susceptible to nucleophilic attack (e.g. displacement of halo-pyridines).

N N N NB: behaves like an imine

Cl Nu Cl Nu Nu Pyridine: lone pair reactivity 111

§ Basicity: - Consider pKa of conjugate acid…

- The stronger the conjugate acid the lower the pKa and the weaker the base.

- The weaker the conjugate acid the higher the pKa and the stronger the base.

- Pyridine lone pair in sp2 orbital (c.f. imine), hence weak base. - Piperidine lone pair in sp3 orbital, hence stronger base (i.e. greater ‘p’ character).

H H H H N N N N + H + H

PYRIDINE PIPERIDINE

pKa 5.2 pKa 11.5

NB: for a recap. of pKa, see ‘OC’ p 197. Question: pyridine basicity

§ Which pyridine is the strongest base?

N Me N Cl N

1 2 3

pKaH 5.2 pKaH 6.8 pKaH 0.7

A. 1 B. 2 C. 3 Pyridine: lone pair reactivity 113

§ Nucleophilicity: - NB: lone pair cannot be delocalised around the ring (i.e. sp2 orbital orthogonal to the ring).

p orbitals

90˚ N N N X!

2 sp orbital

- Hence lone pair is a good nucleophile and readily alkylated or acylated. e.g. O Cl Cl R N R acylated N O

- Typically, the more basic a pyridine, the more 2,6-di-tert-butyl pyridine (good base, poor nucleophile) nucleophilic (except when sterically crowded). N Question: DMAP as a catalyst

§ N,N-Dimethyl-4-amino pyridine (DMAP) is a useful pyridine-based catalyst for acylation reactions. What type of catalysis is this an example of? (hint: what is the pyridine doing and why might DMAP be better than pyridine?) O Ph O O HO Me N Cl 2 + Me2N Cl Ph N Ph Me2N N O N DMAP

A. Acid catalysis B. Base catalysis C. Nucleophilic catalysis Pyridine: benzene-like reactivity 115

§ Electrophilic aromatic substitution: - SEAr readily occurs on benzene, but much less favourable on pyridine. - Electronegative nitrogen atom lowers energy of p orbitals within pyridine ring (w.r.t. benzene), hence pyridine ring less nucleophilic (i.e. lower HOMO) but more electrophilic (i.e. lower LUMO). - Pyridine: nucleophilic lone pair leads to reaction with E+ on nitrogen rather than carbon. e.g.

E N E N nitrogen makes pyridine ring electron

deficient (i.e. less nucleophilic)

- Yields from SEAr reactions are typically poor: e.g. 1. Friedel-Crafts reactions normally fail. N

e.g. 2. Nitration with HNO3 + H2SO4 at 300 ˚C gives < 5% of 3-nitropyridine. NO2

e.g. 3. Halogenation with Cl2 + AlCl3 at 130 ˚C gives only moderate yields of 3-chloropyridine. N

NB: benzene reacts readily under these conditions (see 1st Year lecture notes). Cl Pyridine: benzene-like reactivity 116

§ Electrophilic aromatic substitution: - Electronegative nitrogen atom withdraws electron density; C-2, C-4 and C-6 most affected. e.g.

4 δ+

2 6 N N δ+ δ+ N N N δ-

overall

- C-2 or C-4 substitution results in δ+ localised on nitrogen atom (highly disfavoured). e.g.

E E E E N N N N H H H

- C-3 & C-5 substitution least disfavoured (but still unlikely to occur). e.g. E E E E H H H

N N N N Pyridine: benzene-like reactivity 117

§ Electrophilic aromatic substitution: - Can occur if electron donating group present. e.g.

NO2 HNO3

Me N N H2SO4 2 Me2N N

- If no activating groups present, can convert pyridine into pyridine N-oxide. e.g.

mCPBA NB: can also use H O in AcOH N N 2 2 O

nitrogen lone pair NB: stable solid easily oxidised!

Pyridine N-oxides readily undergo SEAr Pyridine: benzene-like reactivity 118

§ Electrophilic aromatic substitution: - Negative charge on oxygen delocalised into pyridine π-system (i.e. makes ring more electron rich). - Reaction with electrophiles occurs at C-2 or C-4 (but mainly at C-4). e.g. NO2 NO2 H NO2

HNO3 H SO N 2 4 N N O O C-4 attack O Rearomatisation

- Easily cleave N-oxide with P(III) compounds (e.g. PCl3, P(OMe)3) to regain pyridine. e.g. NO2 NO2 O P(OMe)3 + P THF MeO OMe N OMe N O P(OMe)3

Phosphorus acts as a nucleophile (lone pair) Formation of strong P=O

and an electrophile (vacant d orbital) double bond drives reaction

NB: obtain 2-chloropyridine if use PCl3 (for further details see ‘OC’ textbook p 730) Pyridine: imine-like reactivity 119

§ Nucleophilic substitution: - C-2 and C-4 halo-pyridines react easily with nucleophiles (c.f halobenzenes are relatively inert). - C-3 halo-pyridines less reactive (i.e. negative charge cannot be delocalised onto nitrogen). e.g.

Na OMe OMe MeOH N N OMe N Cl Cl

8 NB: 2-chloropyridine is 3 x 10 times more reactive than chlorobenzene

- Pyridine will react with strong nucleophiles (e.g. LiAlH4, Grignard reagents etc.). - - Activated pyridine will react with weaker nucleophiles (e.g. NaBH4, CN). e.g.

H CN N N CN N H H

‘Activated’

(i.e. N-protonated) Pyridine reactivity in action 120

§ Synthesis of Omeprazole: - $Billion-selling drug: proton pump inhibitor to treat stomach ulcers. e.g.

S Ar NO2 Me Me E Me Me Me Me H2O2 HNO3 N Me H SO N Me 2 4 N Me O oxidation nitration O

N-oxide NaOMe SNAr MeOH

H OMe N O Me Me S N steps MeO N Me N Me Me OMe O

Omeprazole Pyridine: summary 121

- Electronic structure of pyridine. - Difference between structure & chemistry of pyridine and benzene. - Basicity: strength relative to non-aromatic amines; effect of substituents. - Nucleophilicity: react through lone pair.

- SEAr: position of substitution; rate of reaction w.r.t. benzene; why unreactive? - Pyridine N-oxides: reactions, methods for synthesis & removal. - Nucleophilic attack: easy with halo-pyridines & activated pyridines. Pyrrole 122

§ Structure: - Isoelectronic with cyclopentadiene anion (i.e. 6π electrons, one in each of 4 parallel p orbitals on the carbons and lone pair in a parallel p orbital on nitrogen). e.g.

sp2 hybridised carbon: H negative charge in p orbital, parallel to plane

of π-system cyclopentadiene one electron in anion each neutral

carbon p orbital

H sp2 hybridised nitrogen: N N H lone pair of electrons in p orbital, parallel to

plane of π-system

pyrrole one electron in each carbon p

orbital Pyrrole vs pyridine 123

§ Structure: - Lone pair on pyridine in sp2 orbital (basic) c.f. lone pair on pyrrole in p orbital (non-basic). e.g.

sp2 hybridised nitrogen: basic N lone pair of electrons in ! N sp2 orbital, orthogonal to plane of π-system

pyridine one electron in each p orbital

H sp2 hybridised nitrogen: N N H lone pair of electrons in p orbital, parallel to non-basic! plane of π-system

pyrrole one electron in each carbon p

orbital

Location of nitrogen lone pair has major effect on reactivity and properties of pyrrole w.r.t pyridine Imidazole: more than one nitrogen in the ring 124

§ Structure: - Imidazole has 6π electrons, hence one nitrogen is pyridine-like and one nitrogen pyrrole-like. e.g.

H N N H N pyrrole-like nitrogen

N non-basic!

imidazole

pyridine-like nitrogen

basic! Pyrrole: NMR spectroscopic properties 125

§ 1H NMR spectroscopic chemical shifts (i.e. δH in ppm): - Pyrrole is electron rich from lone pair donation, hence greater shielding. e.g.

H 7.1 H ≈10 6.5 H 7.5 N H H H 8.5

7.2 N H 6.2

NB: not as big a difference in shifts c.f. pyridine

§ 13C NMR spectroscopic chemical shifts (i.e. δH in ppm): - Pyrrole is electron rich from lone pair donation, hence greater shielding. e.g.

H 123.6

118.0 N 135.7 149.8 128.5 N 107.7 Question: hybridisation states in heterocycles

§ In which hybrid atomic orbital is the nitrogen lone pair located?

H N

A. p B. sp C. sp2 D. sp3 Question: hybridisation states in heterocycles

§ In which hybrid atomic orbital is the nitrogen lone pair located?

H N

A. p B. sp C. sp2 D. sp3 Pyrrole: general reactivity trends 128

§ Lone pair reactivity: - Non-basic (i.e. pKa –3.8), in strong acid pyrrole protonates on carbon not nitrogen. - Nucleophilic on carbon rather than nitrogen (i.e. lone pair delocalised into ring). - Does not undergo N-oxidation (c.f. pyridine).

H H H H E N N N E N H H X! pKa –3.8 H

§ Electrophilic substitution: H H N E N - Electron rich ring is highly reactive to SEAr E (more so than benzene). H

nucleophilic on carbon

§ Nucleophilic substitution: - Electron rich ring unreactive towards nucleophilic attack. - Require electron withdrawing substituent to activate ring.

NB: for an example of SNAr on an activated pyrrole, see ‘OC’ textbook p 738 Pyrrole: lone pair reactivity 129

§ Basicity: - Pyrrole lone pair delocalised around ring, hence non-basic. - c.f. Pyrrolidine lone pair in sp3 orbital, hence stronger base.

H H N N H + H H

PYRROLE

pKa –3.8

H H H N N + H

PYRROLIDINE

pKa 11.0 Pyrrole: lone pair reactivity 130

§ Acidity: - Pyrrole has N-H present (c.f. pyridine). - Weakly acidic (i.e. lone pair delocalised around ring so less electron density located at nitrogen c.f. pyrrolidine). e.g. H N Base N

pKa 16.5

NB: sp2 orbital orthogonal N H N to π-system so no resonance stabilisation

N-H bond uses negative charge in nitrogen sp2 orbital & 2 nitrogen sp orbital lone pair delocalised into aromatic ring

- Pyrrolidine is much less acidic. e.g. H N Base N negative charge in nitrogen sp3 orbital & lone pair fully localised on nitrogen pKa 44.0 Pyrrole: electrophilic substitution 131

§ SEAr: - Pyrrole is electron rich, reacts readily with electrophiles (NB: more reactive than benzene). - Nitrogen lone pair delocalisation increases electron density at each carbon in ring. e.g.

H H H H H H δ+ N N N N N δ- N δ- δ- δ-

overall

- Is substitution at C-2 or C-3 favoured?

H H NB: adding another heteroatom into the ring deactivates N N towards SEAr as this is ‘pyridine-like’ N O Pyrrole: electrophilic substitution 132

§ SEAr at C-2 position: - Generally favoured. - Cation resulting from electrophilic attack is more stabilised (i.e. 3 resonance forms). - Linear conjugated intermediate (i.e. both double bonds conjugated with N+). e.g.

linear conjugated

intermediate

H H H H N E N N N E E E H H H

– H

H N E

C-2 product Pyrrole: electrophilic substitution 133

§ SEAr at C-3 position: - Less favoured. - Cation resulting from electrophilic attack is less stabilised (i.e. 2 resonance forms). - Cross conjugated intermediate (i.e. only one double bond conjugated with N+, less stable than linear conjugated intermediate). e.g. cross conjugated

intermediate H H H N E N N

E H H E E

– H

H N

E

C-3 product Pyrrole: electrophilic substitution 134

e.g. Formylation in the absence of a strong Lewis acid: Vilsmeier-Haack reaction.

H N H NMe2 N O POCl3 + DMF H Cl H Anton Vilsmeier

O O Cl Cl P P O Cl Cl Cl O Cl NMe2 O + P H NMe2 H NMe2 H Cl Cl O

NB: formation of intermediate driven by strong P=O bond formed

NMe2 H H H H H Cl NMe 2 NMe O N N N 2 N H SEAr H H iminium ion H

hydrolysis Aims for this lecture 135

§ Heterocycles: - Continue look at pyrrole reactivity. - Summary & comparison of pyrrole and pyridine structure & reactivity. - Consider epoxide ring opening under acidic and basic conditions. - Synthesis of pyridine and pyrrole. Pyrrole: electrophilic substitution 136

e.g. Recap.: Vilsmeier-Haack reaction.

Anton Vilsmeier

H N H NMe2 N O POCl3 + DMF H Cl H Question: electrophilic substitution of pyrrole

§ What is the product of this Mannich reaction? H N A. 1 H B. 2 NMe2 O + Me2NH ? C. 3 H H H D. 4

H H N N O

H H O 1 2

H H N N

NMe 2 NMe2

3 4 Pyrrole: reaction at nitrogen 138

§ Pyrrole anion undergoes N-alkylation and N-acylation: - Anion in sp2 orbital (90˚ to plane of π-system), so cannot overlap with ring & subsequent reaction occurs at nitrogen. - 2 steps: (1). Deprotonate pyrrole (NB: pKa 16.5), (2) Add electrophile. e.g. 1.

O

O NaH Cl Ph N H N N Ph

N-acylation

NB: without deprotonation would get reaction solely at carbon

e.g. 2.

LDA Me I N H N N Me

N-alkylation Pyrrole: Diels-Alder reactions 139

§ Pyrrole functions as a diene: - NB: benzene does not function as a diene for Diels-Alder reactions (i.e. pyrrole is “less aromatic”). e.g. Synthesis of analgesic compound epibatidine, a natural product.

NB: see Dr Lebrasseur’s course on pericyclic chemistry

Boc Boc N Cl [4+2] steps N N Boc + N

SO2Ar SO2Ar

epibatidine

NB: furan also takes part in Diels-Alder cycloadditions. Pyrrole: summary 140

- Electronic structure of pyrrole. - Difference between structure & chemistry of pyrrole w.r.t. pyridine and benzene. - Recognise pyrrole-like nitrogens and pyridine-like nitrogens in heterocycles. - Basicity: pyrrole is not basic; protonate at C-2 in strong acid (leads to polymerisation). - Acidity: pyrrole is weakly acidic; comparison to non-aromatic amines; electrophilic attack at nitrogen through pyrrole anion. - Nucleophilicity: nitrogen lone pair delocalisation makes pyrrole ring electron rich; high nucleophilicity (at carbon). - SEAr: position of substitution (i.e. C-2); rate of reaction w.r.t. benzene; why more reactive? - Vilsmeier-Haack reaction; Mannich reaction. - Nucleophilic attack: unreactive, require electron withdrawing substituents to activate pyrrole ring. Pyrrole & pyridine: schematic reactivity summary 141

E+ (easy) E+ (difficult)

Base Base (difficult) (difficult) H N H Nu- Nu- (difficult) (easy) BENZENE PYRIDINE

E+ (very easy)

N N H Base (very easy) N E+ (very easy) Nu- (very H easy)

PYRROLE PYRROLE ANION ACTIVATED PYRIDINE Useful reactions of another heterocycle: epoxides 142

§ Epoxide ring-opening: - Strained 3-membered ring, but requires either a good nucleophile or acid catalysis to react well.

- SN2 mechanism, hence inversion of stereochemistry at reaction centre (i.e. stereospecific). e.g.

OH 1). Nu NB: inversion of * stereochemistry at O 2). H3O reacting carbon centre Nu SN2

- Unsymmetrical epoxides lead to issues of regioselectivity (i.e. which end reacts?). e.g.

OH Nu 1). Nu or O 2). H3O Nu OH

attack at most attack at less

hindered end hindered end Useful reactions of other heterocycles 143

§ Epoxide ring-opening in base: - Attack less hindered end of epoxide (i.e. minimise steric interactions between Nu- and E+).

- Pure SN2 mechanism, inversion of stereochemistry at reaction centre (i.e. stereospecific). e.g.

NB: inversion

OH OMe Na OMe or * O MeOH OMe OH

NB: epoxide oxygen is a poor LG (i.e. RO-), attack at less - so needs a strong Nu hindered end

Pentacoordinate T.S. in SN2 reaction hence minimise steric interactions

NB: for further details on epoxide ring opening, see ‘OC’ textbook p 438 Useful reactions of other heterocycles 144

§ Epoxide ring-opening in acid: - Attack more hindered end of epoxide (i.e. epoxide oxygen protonated in acid, so build up of postive charge in T.S. stabilised at most substituted end). - ‘Loose’ SN2 transition state, inversion of stereochemistry at reaction centre (i.e. stereospecific). e.g. NB: inversion

OH OMe MeOH, HCl * or O H OMe OH

NB: epoxide oxygen is protonated, attack at more

so it is a good LG (i.e. ROH) and hindered end does not need a strong Nu-

MeOH H (+) H OMe OMe

(+) (+) O (+) O (+) O O H H H H H H H

attack site best able to loose SN2 transition state stabilise partial +ve charge

NB: for further details on epoxide ring opening, see ‘OC’ textbook p 438 Question: epoxide opening

§ Identify the correct product from this epoxide ring opening.

A. 1 MeOH, HCl O ? B. 2 C. 3 D. 4

OH OMe

OMe OH 1 2

OH OMe

OMe OH 3 4 Synthesis of heterocyclic rings 146

§ In general: - Based on simple carbonyl chemistry (see previous recap. of key reactions). - Can disconnect the main bonds to reveal simple linear precursors. e.g. Pyrrole

R R R R + N HN R R R'NH2 O O O R' R'

pyrrole 1,4-dicarbonyl compound

e.g. Pyridine

+ NH3 R R R N R R R O H2N O O

pyridine 1,5-dicarbonyl compound

NB: for a further summary of 5- and 6-membered ring synthesis, see ‘OC’ textbook p 785 & 786. Synthesis of pyrroles 147

§ Paal-Knorr synthesis: - Uses a 1,4-dicarbonyl compound and either ammonia or a 1˚ amine. - Requires weak acid (NB: strong acid will lead to formation of the corresponding furan). e.g.

R R AcOH Ludwig Knorr 1,4-dicarbonyl O O R R compound N R'NH 2 R = H, alkyl, aryl R' pyrrole - Mechanism: e.g. NB: enamine H formation NH2 O OH AcOH cyclisation NH O N H Cl H O Ar Ar

± H H

OH2 N N N Cl Ar Ar hemiaminal NB: overall, lose 2 moles of water Question: synthesis of pyrroles

§ Identify the correct pyrrole product from this reaction

A. 1 O H N 2 AcOH B. 2 + ? C. 3 O D. 4

N

1 2

N

3 4

N N Synthesis of pyridines 149

§ Hantzsch synthesis: - 4 component reaction (NB: 3 different substrates). - Initially affords a 1,4-dihydropyridine, but readily oxidises in air to give the pyridine. e.g.

Arthur Rudolf Hantzsch

R R R O H pH 8.5 EtO2C CO2Et [O] EtO C CO Et EtO2C CO2Et 2 2 EtOH oxidation N O O H N NH3 1,4-dihydropyridine pyridine

NB: aldehyde provides extra carbon for pyridine ring (c.f. pyrrole synthesis)

NB: for further details of the Hantzsch pyridine ring synthesis, see ‘OC’ textbook p 763. Synthesis of pyridines 150

§ Mechanism: - Condensation with aldehyde to form α,β-unsaturated product (c.f. aldol reaction). R O OH R O H E1 EtO2C EtO2C ± H EtO2C cb EtO2C R R

O O O O

α,β-unsaturated product

- α,β-Unsaturated product is good conjugate acceptor, so Michael reaction occurs.

NB: 2nd equivalent of enolate used NB: enamine formation R R R EtO2C CO2Et EtO2C CO2Et EtO2C CO2Et

O O O O O O

conjugate addition 1,5-dicarbonyl compound

NB: for further details of the Hantzsch pyridine ring synthesis, see ‘OC’ textbook p 763. Synthesis of pyridines 151

§ Mechanism (continued): - Enamine formation, then intramolecular cyclisation of nitrogen onto other ketone.

R R R R H EtO2C CO2Et EtO2C CO2Et EtO2C CO2Et EtO2C CO2Et NH3

O O O N N H2N H H enamine formation cyclisation 1,4-dihydropyridine

- Final oxidation step is facile. - Occurs in air, or with chemical reagents such as DDQ.

R R O EtO2C CO2Et oxidation EtO2C CO2Et Cl CN

N H N Cl CN O

Dichlorodicyanoquinone NB: The mechanism for oxidation is beyond the scope of this course, (DDQ) but for further details see ‘OC’ textbook p 764 Summary of the lecture 152

- Under acidic conditions, epoxides open at the more hindered end (i.e. loose SN2 T.S.). - Under basic conditions, epoxides open at the less hindered end (i.e. minimise steric crowding in T.S.). - Synthesise pyrroles and pyridines using standard carbonyl chemistry (e.g. aldol reaction, conjugate addition, imine & enamine formation etc.). - Can use Paal-Knorr pyrrole synthesis to make pyrroles from 1,4-dicarbonyl compounds. - Can use Hantzsch pyridine synthesis to make pyridines via a 4 component coupling reaction. Exam Questions 153

§ Representative questions

This a new course and hence there is only one mock exam paper (see QMplus) that relates specifically to this course (and the exclusion of any other). However, much of the material is what I would class as core material that will crop up across a range of examination questions. With regards to older lecture courses, particular attention should be paid to parts of the SBC703 course, ‘Synthesis of Pharmaceutically Active Molecules’. Access to past papers can be made via the library webpage. For general practice, many textbooks contain questions that will relate to the topics covered.